<p>Let \(f(x) = \max\left\{\sin x,\, \cos x,\, \dfrac{1}{2}\right\}\) then determine the area of region bounded by the curves \(y = f(x)\), \(x\)-axis, \(y\)-axis and \(x = 2\pi\).</p>
<p>\(\left(\dfrac{5\pi}{12} - \sqrt{2} + \sqrt{3}\right)\)</p>
<p>\(\left(\dfrac{5\pi}{12} + \sqrt{2} + \sqrt{3}\right)\)</p>
<p>\(\left(\dfrac{5\pi}{12} + \sqrt{2} - \sqrt{3}\right)\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: The function f(x) = max{sin x, cos x, 1/2} requires identifying three distinct regions where each component dominates, then integrating piecewise over [0, 2π]. Key intersections occur where sin x = 1/2, cos x = 1/2, and sin x = cos x.
<p><strong>Step 1: Identify critical intersection points in [0, 2π]</strong></p><p>• sin x = cos x: x = π/4, 5π/4</p><p>• sin x = 1/2: x = π/6, 5π/6</p><p>• cos x = 1/2: x = π/3, 5π/3</p><p><strong>Step 2: Determine which function is maximum in each interval</strong></p><p>• [0, π/6]: cos x ≥ sin x and cos x ≥ 1/2, so f(x) = cos x</p><p>• [π/6, π/4]: cos x is max</p><p>• [π/4, π/3]: sin x ≥ cos x but need to check against 1/2; sin x is max</p><p>• [π/3, 5π/6]: sin x ≥ 1/2, so f(x) = sin x</p><p>• [5π/6, 5π/4]: 1/2 ≥ both sin x and cos x, so f(x) = 1/2</p><p>• [5π/4, 5π/3]: cos x is max (negative then positive)</p><p>• [5π/3, 2π]: cos x ≥ 1/2, so f(x) = cos x</p><p><strong>Step 3: Calculate the area</strong></p><p>Area = ∫₀^(π/3) cos x dx + ∫_(π/3)^(5π/6) sin x dx + ∫_(5π/6)^(5π/4) (1/2) dx + ∫_(5π/4)^(5π/3) cos x dx + ∫_(5π/3)^(2π) cos x dx</p><p>= [sin x]₀^(π/3) + [-cos x]_(π/3)^(5π/6) + [(1/2)x]_(5π/6)^(5π/4) + [sin x]_(5π/4)^(5π/3) + [sin x]_(5π/3)^(2π)</p><p>= (√3/2) + (√3/2 + 1/2) + (5π/8 - 5π/12) + (-√3/2 + √2/2) + (√3/2)</p><p>= √3 + 1/2 + 5π/24 + √2/2</p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B